Question

The figure shows a region of length ${}^{\prime }{l}^{\prime }$ with a uniform magnetic field of $0.3\text{T}$ in it and a proton entering the region with velocity $4\times {10}^5{\text{ms}}^{-1}$ making an angle ${60}^{\circ }$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, ${}^{\prime }{l}^{\prime }$ is close to (mass of proton $=1.67\times {10}^{-27}\text{kg}$, charge of the proton = $1.6\times {10}^{-19}\text{C}$)

• A. $0.11\text{m}$
• B. $0.88\text{m}$
• C. $0.44\text{m}$
• D. $0.22\text{m}$

Answer 1

The correct option is C $0.44\text{m}$

Time period of one revolution of portion, $T=\frac{2\pi m}{qB}$

Here, $m=$ mass of proton
$q=$ charge of proton
$B=$ magnetic field

Linear distance travelled in one revolution,
$p=T(v\cos \theta$)
(Here, $v=$ velocity of proton)
$\therefore$ Length of region,

$\Rightarrow l=10\times v\cos {60}^o\times \frac{2\pi m}{qB}$

$\Rightarrow l=\frac{10\times 3.14\times 1.67\times {10}^{-27}\times 4\times {10}^5}{1.6\times {10}^{-19}\times 0.3}$

$\Rightarrow l=0.44\text{m}$